 taking this alkene and reacting it with bromine and hydrogen sulfide, notice it says excess hydrogen sulfide and DCM. Now, DCM, that's just the solvent, so it's not going to do participate anything in the reaction. So if you want to, you can just kind of get rid of it. In fact, DCM, you'll see it written like DCM or CH2, Cl2, of course that's the formula for DCM. But anyway, so the first step of this reaction is going to take bromine and remember when you have just bromine and an alkene, you're going to make the bromonium ion. So this does a similar thing, we actually make that bromonium ion. Great soft thing. And remember the bromonium ion is this three member ring. And if you can't remember, does it have a charge? Does it not have a charge? Remember both sides of the equation have to have the same amount of charge. So since you started with zero charge over here, bromine has zero, valcine has zero, and you're making negative charge, it's an easy way to remember the bromonium ion has to have that positive charge. Anyways, now you're going to use the hydrogen sulfide. It's going to act as a nucleophile. Similar to what water would and the reaction of making the halo hydrant. So this is essentially an analogy of that halo hydrogen reaction. So the bromonium ion is very electrophilic at these two carbons. So you're going to have this SN2 reaction. So this is going to be a good nucleophile, this is a good electrophile of course. But you have the choice of which carbon to react. Of course, this carbon here is very sterically hindered. This carbon here is less sterically hindered, so it's only going to react on this side. And of course, you're going to make the enantiomer of that bromonium ion too. So I'm going to get this reaction here, the sulfur with that positive charge pointing backwards. And then it says it's got excess hydrogen sulfide that's of course the deprotonate. So you're going to get that plus it's an enantiomer. Why the enantiomer? Because this here and this here are sp2 carbons. So the bromine can attack from this front or top face or the back face like that. So you'll get an equal amount of the enantiomers here. And of course, then that's going to give you that. So what do you have? You've got the making of the bromonium ion, the SN2 reaction here, and the acid base or deprotonation that gives you the final product. Cool. Any questions?